Discussion on Get Node Value Challenge

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1 week ago+ 0 comments

def getNode(llist, positionFromTail):
    # Write your code here
    last_index = llist
    
    for _ in range(positionFromTail):
        last_index = last_index.next

    desire_position = llist
    
    while last_index.next:
        last_index = last_index.next
        desire_position = desire_position.next
        
    return desire_position.data

2 weeks ago+ 0 comments

here's my C++ solution. If we can get the length of the linked list, we can get that position form the head. Why? Because I'm initially given the head pointer so I just have to count certain number from that head.

int getNode(SinglyLinkedListNode* llist, int positionFromTail)
{
    int LSize = 0;
    SinglyLinkedListNode* counter = llist;
    while(counter)
    {
        ++LSize;
        counter = counter->next;
    }
    int positionFromHead = LSize - positionFromTail;
    while (positionFromHead-- > 1)
    {
        llist = llist->next;
    }
    return llist->data;
}

2 weeks ago+ 0 comments

Javascript Solution using Recursion. Time complexity O(N). No Arrays

function getNode(llist, positionFromTail) {

    let counter = 0;
    let flag = false;
    let res = null;
    function recur(llist,positionFromTail,flag){
        if(llist && !flag){
            recur(llist.next,positionFromTail,flag)
            counter += 1;
            if(counter == positionFromTail+1) {
                res = llist.data;
                flag = true;
            }
        } 
    }
    recur(llist,positionFromTail)
    return res;

}

2 weeks ago+ 0 comments

Python3 soln:

def getNode(llist, positionFromTail):
        """
        find length of list
        position from head = length - position from tail
        iterate again, if traversals = position, print value
        """
        length, traversals = 0,0
        current = llist
        while current:
                length += 1
                current = current.next

        positionFromHead = length-positionFromTail-1
        for i in range(positionFromHead):
                llist = llist.next

return llist.data

4 weeks ago+ 0 comments

javascript code:

function getNode(llist, positionFromTail) {
    let data;
    let curr = llist
    for(let i = 0;i<positionFromTail;i++){
        curr = curr.next;
    }
    if(curr.next){
        llist = llist.next
        while(curr.next){
            data = llist.data;
            curr = curr.next;
            llist = llist.next
        }
        return data
    }
    return llist.data
}

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